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Find Intersection of all Intervals
• Difficulty Level : Medium
• Last Updated : 26 Feb, 2019

Given N intervals of the form of [l, r], the task is to find the intersection of all the intervals. An intersection is an interval that lies within all of the given intervals. If no such intersection exists then print -1.

Examples:

Input: arr[] = {{1, 6}, {2, 8}, {3, 10}, {5, 8}}
Output: [5, 6]
[5, 6] is the common interval that lies in all the given intervals.

Input: arr[] = {{1, 6}, {8, 18}}
Output: -1
No intersection exists between the two given ranges.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:

• Start by considering first interval as the required answer.
• Now, starting from the second interval, try searching for the intersection. Two cases can arise:
1. There exists no intersection between [l1, r1] and [l2, r2]. Possible only when r1 < l2 or r2 < l1. In such a case answer will be 0 i.e. no intersection exists.
2. There exists an intersection between [l1, r1] and [l2, r2]. Then the required intersection will be [max(l1, l2), min(r1, r2)].

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;`` ` `// Function to print the intersection``void` `findIntersection(``int` `intervals[][2], ``int` `N)``{``    ``// First interval``    ``int` `l = intervals[0][0];``    ``int` `r = intervals[0][1];`` ` `    ``// Check rest of the intervals and find the intersection``    ``for` `(``int` `i = 1; i < N; i++) {`` ` `        ``// If no intersection exists``        ``if` `(intervals[i][0] > r || intervals[i][1] < l) {``            ``cout << -1;``            ``return``;``        ``}`` ` `        ``// Else update the intersection``        ``else` `{``            ``l = max(l, intervals[i][0]);``            ``r = min(r, intervals[i][1]);``        ``}``    ``}`` ` `    ``cout << ``"["` `<< l << ``", "` `<< r << ``"]"``;``}`` ` `// Driver code``int` `main()``{``    ``int` `intervals[][2] = {``        ``{ 1, 6 },``        ``{ 2, 8 },``        ``{ 3, 10 },``        ``{ 5, 8 }``    ``};``    ``int` `N = ``sizeof``(intervals) / ``sizeof``(intervals[0]);``    ``findIntersection(intervals, N);``}`

## Java

 `// Java implementation of the approach``import` `java.io.*;`` ` `class` `GFG``{``     ` `// Function to print the intersection``static` `void` `findIntersection(``int` `intervals[][], ``int` `N)``{``    ``// First interval``    ``int` `l = intervals[``0``][``0``];``    ``int` `r = intervals[``0``][``1``];`` ` `    ``// Check rest of the intervals``    ``// and find the intersection``    ``for` `(``int` `i = ``1``; i < N; i++) ``    ``{`` ` `        ``// If no intersection exists``        ``if` `(intervals[i][``0``] > r || ``            ``intervals[i][``1``] < l) ``        ``{``            ``System.out.println(-``1``);``            ``return``;``        ``}`` ` `        ``// Else update the intersection``        ``else``        ``{``            ``l = Math.max(l, intervals[i][``0``]);``            ``r = Math.min(r, intervals[i][``1``]);``        ``}``    ``}``    ``System.out.println (``"["` `+ l +``", "` `+ r + ``"]"``);``}`` ` `    ``// Driver code``    ``public` `static` `void` `main (String[] args) ``    ``{`` ` `        ``int` `intervals[][] = {{ ``1``, ``6` `},``                            ``{ ``2``, ``8` `},``                            ``{ ``3``, ``10` `},``                            ``{ ``5``, ``8` `}};``        ``int` `N = intervals.length;``        ``findIntersection(intervals, N);``    ``}``}`` ` `// This Code is contributed by ajit.. `

## Python

 `# Python3 implementation of the approach`` ` `# Function to print the intersection``def` `findIntersection(intervals,N):`` ` `    ``# First interval``    ``l ``=` `intervals[``0``][``0``]``    ``r ``=` `intervals[``0``][``1``]`` ` `    ``# Check rest of the intervals ``    ``# and find the intersection``    ``for` `i ``in` `range``(``1``,N):`` ` `        ``# If no intersection exists``        ``if` `(intervals[i][``0``] > r ``or` `intervals[i][``1``] < l):``            ``print``(``-``1``)`` ` `        ``# Else update the intersection``        ``else``:``            ``l ``=` `max``(l, intervals[i][``0``])``            ``r ``=` `min``(r, intervals[i][``1``])``         ` `     ` ` ` `    ``print``(``"["``,l,``", "``,r,``"]"``)`` ` `# Driver code`` ` `intervals``=` `[``            ``[ ``1``, ``6` `],``            ``[ ``2``, ``8` `],``            ``[ ``3``, ``10` `],``            ``[ ``5``, ``8` `]``            ``]``N ``=``len``(intervals)``findIntersection(intervals, N)`` ` `# this code is contributed by mohit kumar`

## C#

 `// C# implementation of the approach ``using` `System;`` ` `class` `GFG ``{ ``     ` `// Function to print the intersection ``static` `void` `findIntersection(``int` `[,]intervals, ``int` `N) ``{ ``    ``// First interval ``    ``int` `l = intervals[0, 0]; ``    ``int` `r = intervals[0, 1]; `` ` `    ``// Check rest of the intervals ``    ``// and find the intersection ``    ``for` `(``int` `i = 1; i < N; i++) ``    ``{ `` ` `        ``// If no intersection exists ``        ``if` `(intervals[i, 0] > r || ``            ``intervals[i, 1] < l) ``        ``{ ``            ``Console.WriteLine(-1); ``            ``return``; ``        ``} `` ` `        ``// Else update the intersection ``        ``else``        ``{ ``            ``l = Math.Max(l, intervals[i, 0]); ``            ``r = Math.Min(r, intervals[i, 1]); ``        ``} ``    ``} ``    ``Console.WriteLine(``"["` `+ l + ``", "` `+ r + ``"]"``); ``} `` ` `// Driver code ``public` `static` `void` `Main() ``{ ``    ``int` `[,]intervals = {{ 1, 6 }, { 2, 8 }, ``                        ``{ 3, 10 }, { 5, 8 }}; ``    ``int` `N = intervals.GetLength(0); ``    ``findIntersection(intervals, N); ``} ``} `` ` `// This code is contributed by Ryuga`

## PHP

 ` ``\$r` `|| ``            ``\$intervals``[``\$i``][1] < ``\$l``) ``        ``{``            ``echo` `-1;``            ``return``;``        ``}`` ` `        ``// Else update the intersection``        ``else``        ``{``            ``\$l` `= max(``\$l``, ``\$intervals``[``\$i``][0]);``            ``\$r` `= min(``\$r``, ``\$intervals``[``\$i``][1]);``        ``}``    ``}`` ` `    ``echo` `"["` `. ``\$l` `. ``", "` `. ``\$r` `. ``"]"``;``}`` ` `// Driver code``\$intervals` `= ``array``(``array``(1, 6), ``array``(2, 8),``                   ``array``(3, 10), ``array``(5, 8));``\$N` `= sizeof(``\$intervals``);``findIntersection(``\$intervals``, ``\$N``);`` ` `// This code is contributed``// by Akanksha Rai``?>`
Output:
```[5, 6]
```

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